TOPICS

Exercise - 1.5

Number Systems

**Question-1 :-** Classify the following numbers as rational or irrational:

(i) 2 - √5, (ii) (3+√23) - √23, (iii) 2√7/7√7, (iv) 1/√2, (v) 2π

(i) 2 - √5 = 2 - 2.236067... = 0.236067 As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number. (ii) (3+√23) - √23 = 3+√23-√23 = 3 =3/1 therefore, it is a rational no. and it can be represented in p/q. (iii) 2√7/7√7 = 2/7 therefore, it is a rational no. and it can be represented in p/q. (iv) 1/√2 = 1/√2 x √2/√2 = √2/2 = 0.7071067811... As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number. (v) 2π = 2 x 3.1415... = 6.2830... As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number.

**Question-2 :-** Simplify each of the following expressions:

(i) (3+√3)(2+√2), (ii) (3+√3)(3-√3), (iii) (√5+√2)², (iv) (√5-√2)(√5+√2)

(i) (3+√3)(2+√2) =3(2+√2)+√3(2+√2) =6+3√2+2√3+√6 (ii) (3+√3)(3-√3) =3²-(√3)² =9-3 = 6 (iii) (√5+√2)² =(√5)²+(√2)²+2√5√2 =5+2+2√10 =7+2√10 (iv) (√5-√2)(√5+√2) =(√5)²-(√2)² =5-2 = 3

**Question-3 :-** Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is,
π = c/d This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

There is no contradiction. When we measure a length with scale or any other instrument, we only obtain an approximate rational value.We never obtain an exact value. For this reason, we may not realise that either c or d is an irrational. Therefore, the c/d fraction is irrational. Hence, π is an irrational.

**Question-4 :-** Represent √9.3 . on the number line.

Mark a line segment OB = 9.3 on number line. Further, take BC of 1 unit. Find the midpoint D of OC and draw a semi-circle on OC while taking D as its centre. Draw a perpendicular to line OC passing through point B. Let it intersect the semi-circle at E. Taking B as centre and BE as radius, draw an arc intersecting number line at F. BF is √9.3

**Question-5 :-** Rationalise the denominators of the following:

(i) 1/√7, (ii) 1/(√7-√6), (iii) 1/(√5+√2), (iv) 1/(√7-2)

(i) 1/√7 Here 1 is the Numenator and √7 is Denominator, so multiply and divide by √7 =1/√7 x √7/√7 =√7/7 (ii) 1/(√7-√6) Here 1 is the Numenator and (√7-√6) is Denominator, so multiply and divide by (√7+√6) =1/(√7-√6) x (√7+√6)/(√7+√6) =(√7+√6)/(7-6) =(√7+√6)/1 = √7+√6 (iii) 1/(√5+√2) Here 1 is the Numenator and (√5-√2) is Denominator, so multiply and divide by (√5-√2) =1/(√5+√6) x (√5-√2)/(√5-√2) =(√5-√2)/(5-2) =(√5-√2)/3 (iv) 1/(√7-2) Here 1 is the Numenator and (√7-2) is Denominator, so multiply and divide by (√7+2) =1/(√7-2) x (√7+2)/(√7+2) =(√7+2)/(7-4) =(√7+2)/3

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